So is it safe to say that we can use integration by substitution???
I would simplify the integrand by factoring $\displaystyle \sqrt{t}$ from the numerator, then dividing out the common factors of the numerator and denominator. Or you might find it easier to just split the fraction, then reduce each term.
You can do this with the help of integration by parts.
Now after simplification:
$\displaystyle \int \frac{t + 2t^2}{\sqrt{t}}\;\; dt = \int (\sqrt{t} + 2t\sqrt{t})\;\;dt = \int \sqrt{t}\;\;dt + \int 2t\sqrt{t}\;\;dt$
We divide this integration in two parts. First:
$\displaystyle \int \sqrt{t}\;\;dt = \frac{2}{3}t^{\frac{3}{2}} + C_1$
Then for $\displaystyle \int 2t\sqrt{t}\;\; dt$ we use integration by parts:
$\displaystyle \text{Let } u = t \;\;\therefore du = dt$
$\displaystyle \text{Let } dv = 2\sqrt{t}\;\; dt\;\; \therefore v = \frac{4}{3}t^{\frac{3}{2}}$
Now:
$\displaystyle \begin{align*}\int 2t\sqrt{t}\;\;dt =& \frac{4}{3}t^{\frac{5}{2}} - \int \frac{4}{3}t^{\frac{3}{2}}\;\; dt \\ =& \frac{4}{3}t^{\frac{5}{2}} - \frac{8}{15}t^{\frac{5}{2}} + C_2 \\ =& \frac{4}{5}t^{\frac{5}{2}}+ C_2...........\text{[After simplicfication]}\end{align*}$
So:
$\displaystyle \therefore \int \frac{t + 2t^2}{\sqrt{t}}\;\; dt = \frac{2}{3}t^{\frac{3}{2}} + \frac{4}{5}t^{\frac{5}{2}} + C$
Check:
integration (t + 2*t^2)/(t)^(1/2) dt - Wolfram|Alpha